9. -/1 points a 4.4 - g bullet leaves the muzzle of a rifle with a speed of 330 m/s. what force (assumed constant) is exerted on the bullet while it is traveling down the 0.77 - m - long barrel of the rifle? n resources read it

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# Explanation:
## Step1: Convert mass to SI unit
The mass of the bullet $m = 4.4\ g=4.4\times10^{- 3}\ kg$.
## Step2: Use the kinematic - equation $v^{2}=v_{0}^{2}+2ax$
The bullet starts from rest, so $v_{0} = 0\ m/s$, $v = 330\ m/s$ and $x = 0.77\ m$.
We can solve for the acceleration $a$:
$$
\begin{align*}
v^{2}&=v_{0}^{2}+2ax\
330^{2}&=0 + 2a\times0.77\
a&=\frac{330^{2}}{2\times0.77}\
a&=\frac{108900}{1.54}\
a& = 70714.29\ m/s^{2}
\end{align*}
$$
## Step3: Use Newton's second - law $F = ma$
$$
\begin{align*}
F&=m\times a\
F&=(4.4\times 10^{-3})\times70714.29\
F&=311.14\ N
\end{align*}
$$
# Answer:
$311.14$

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Question

Explanation:

Step1: Convert mass to SI unit

Step2: Use the kinematic - equation $v^{2}=v_{0}^{2}+2ax$

Step3: Use Newton's second - law $F = ma$

Answer: